An electron has a total energy of 2 MeV. Calculate the effective mass of the electron in kg and its speed. Assume rest mass of electron 0.511 MeV.

Answer

To solve the problem, we need to calculate the effective mass of the electron and its speed given the total energy and the rest mass of the electron.

### Step 1: Calculate the Effective Mass of the Electron

The effective mass \( m_{\text{eff}} \) of the electron can be calculated using the formula:

\[ m_{\text{eff}} = m_0 \cdot \frac{E}{E_0} \]

where:
- \( m_0 \) is the rest mass of the electron (0.511 MeV),
- \( E \) is the total energy of the electron (2 MeV),
- \( E_0 \) is the rest energy of the electron (0.511 MeV).

Substituting the values:

\[ m_{\text{eff}} = (9.1 \times 10^{-31} \text{ kg}) \cdot \frac{2 \text{ MeV}}{0.511 \text{ MeV}} \]

Calculating the fraction:

\[ \frac{2}{0.511} \approx 3.91 \]

Now, substituting this back into the equation:

\[ m_{\text{eff}} = (9.1 \times 10^{-31} \text{ kg}) \cdot 3.91 \approx 3.56 \times 10^{-30} \text{ kg} \]

### Step 2: Calculate the Speed of the Electron

To find the speed \( u \) of the electron, we can use the relation between the effective mass and speed:

\[ m_{\text{eff}} = \frac{m_0}{\sqrt{1 - \left(\frac{u}{c}\right)^2}} \]

Rearranging the formula gives:

\[ \sqrt{1 - \left(\frac{u}{c}\right)^2} = \frac{m_0}{m_{\text{eff}}} \]

Squaring both sides:

\[ 1 - \left(\frac{u}{c}\right)^2 = \left(\frac{m_0}{m_{\text{eff}}}\right)^2 \]

Now, substituting the values:

\[ \left(\frac{u}{c}\right)^2 = 1 - \left(\frac{9.1 \times 10^{-31}}{3.56 \times 10^{-30}}\right)^2 \]

Calculating the fraction:

\[ \frac{9.1 \times 10^{-31}}{3.56 \times 10^{-30}} \approx 0.255 \]

Now squaring it:

\[ \left(0.255\right)^2 \approx 0.065 \]

Now substituting back:

\[ \left(\frac{u}{c}\right)^2 = 1 - 0.065 \approx 0.935 \]

Taking the square root:

\[ \frac{u}{c} \approx \sqrt{0.935} \approx 0.967 \]

Now, multiplying by the speed of light \( c = 3 \times 10^8 \text{ m/s} \):

\[ u \approx 0.967 \times 3 \times 10^8 \approx 2.9 \times 10^8 \text{ m/s} \]

### Final Answers

- Effective mass of the electron: \( m_{\text{eff}} \approx 3.56 \times 10^{-30} \text{ kg} \)
- Speed of the electron: \( u \approx 2.9 \times 10^8 \text{ m/s} \)